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\(A=\frac{x}{xy+x+xyz}+\frac{y}{yz+y+1}+\frac{z}{xz+z+xyz}\)
\(=\frac{1+y+yz}{y+yz+1}=1\)
TA CÓ \(\frac{x}{xy+x+1}\)+\(\frac{y}{yz+y+1}\)+\(\frac{z}{xz+z+1}\)
=\(\frac{x}{xy+x+1}\)+\(\frac{xy}{xyz+xy+x}\)+\(\frac{xyz}{x^2yz+xyz+xy}\)
=\(\frac{x}{xy+x+1}\)+\(\frac{xy}{xy+x+1}\)+\(\frac{1}{xy+x+1}\)(vì xyz=1)
=\(\frac{x+xy+1}{xy+x+1}\)
= 1
Vì xyz = 1 nên x = y = z = 1
=> \(A=\frac{1}{1.1+1+1}+\frac{1}{1.1+1+1}+\frac{1}{1.1+1+1}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)
\(A=\frac{x}{xy+x+1}+\frac{y}{y+1+yz}+\frac{z}{1+z+xz}\)
\(=\frac{x}{xy+x+1}+\frac{xy}{xy+x+xyz}+\frac{xyz}{xy+xyz+x^2yz}\)
\(=\frac{x}{xy+x+1}+\frac{xy}{xy+x+1}+\frac{1}{xy+1+x}\)
\(=\frac{xy+x+1}{xy+x+1}=1\)
\(\frac{x}{xy+x+1}+\frac{xy}{yx+x+xyz}+\frac{xyz}{xy+xyz+x^2yz}\)
\(\frac{x}{xy+x+1}+\frac{xy}{yx+x+1}+\frac{1}{xy+1+x}\)
\(\frac{x+xy+1}{xy+x+1}=1\)
Ta có: \(A=\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1}\)
\(A=\frac{xz}{xyz+xz+z}+\frac{xyz}{xyz^2+xyz+xz}+\frac{z}{xz+z+1}\)
\(A=\frac{xz}{1+xz+z}+\frac{xyz}{z+1+xz}+\frac{z}{xz+z+1}\)
\(A=\frac{xyz+xz+1}{xyz+xz+1}\)
\(A=1\)
Vậy \(A=1\)
Ta có:\(\frac{x}{xy+x+1}=\frac{y}{yz+y+1}=\frac{z}{xz+x+1}\)=\(\frac{xz}{xyz+xz+z}=\frac{yxz}{xyz^2+yxz+xz}=\frac{z}{xz+z+1}\)
=\(\frac{xz}{1+xz+z}=\frac{xyz}{z+1+xz}=\frac{z}{xz+z+1}\)
=\(\frac{xyz+xz+1}{xyz+xz+1}\)=1
Đề bn ghi sai nha~~
Ta có \(\hept{\begin{cases}xyz=1\\A=\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1}\end{cases}}\)
\(\Leftrightarrow A=\frac{x}{xy+x+1}+\frac{yx}{xyz+xy+x}+\frac{xyz}{x^2yz+xyz+xy}\)
\(\Leftrightarrow A=\frac{x}{xy+x+1}+\frac{yx}{x+xy+1}+\frac{1}{x+1+xy}\)
\(\Leftrightarrow A=\frac{x+xy+1}{xy+x+1}=1\)
Vậy A = 1
~~~ Học tốt
Takigawa Miraii
Cách khác
Ta có \(\hept{\begin{cases}xyz=1\\A=\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1}\end{cases}}\)
<=> \(A=\frac{xyz}{xy^2z+xyz+yz}+\frac{y}{yz+y+1}+\frac{zy}{xyz+zy+y}\)
<=> \(A=\frac{1}{y+1+yz}+\frac{y}{yz+y+1}+\frac{zy}{1+zy+y}\)
<=> \(A=\frac{1+y+zy}{y+1+yz}=1\)
Vậy A = 1
Còn 4 cách nx